Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100000, M = 30;
int n, x[N], y[N], xx[N], yy[N], b[M], u[N];
void xo(int *x, vector<pair<int, int>> &rx, int l, int r) {
  rx.push_back({l + 1, r + 1});
  x[l] ^= x[r];
}
void sw(int *x, vector<pair<int, int>> &rx, int l, int r) {
  if (l != r) {
    xo(x, rx, l, r);
    xo(x, rx, r, l);
    xo(x, rx, l, r);
  }
}
int BT(int a, int p) { return bool(a & (1 << p)); }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &x[i]), xx[i] = x[i];
  for (int i = 0; i < n; ++i) scanf("%d", &y[i]), yy[i] = y[i];
  vector<pair<int, int>> rx, ry;
  for (int i = 0; i < M; ++i) {
    int m = -1, bi = 1 << i;
    for (int j = 0; j < n; ++j)
      if (!u[j] && (x[j] & bi)) m = j;
    b[i] = m;
    if (m == -1) continue;
    u[m] = 1;
    for (int j = 0; j < n; ++j)
      if ((x[j] & bi) && (j != m)) xo(x, rx, j, m);
  }
  for (int i = 0; i < n; ++i) {
    int t = y[i];
    for (int j = 0; j < M; ++j)
      if (t & (1 << j)) {
        if (b[j] == -1) {
          puts("-1");
          return 0;
        }
        t ^= x[b[j]];
      }
    if (t) throw;
  }
  for (int i = 0; i < n; ++i) u[i] = 0;
  for (int i = 0; i < M; ++i) {
    int m = -1, bi = 1 << i;
    for (int j = 0; j < n; ++j)
      if (!u[j] && (y[j] & bi)) m = j;
    if (m == -1) {
      if (b[i] != -1) xo(x, rx, b[i], b[i]);
      continue;
    }
    if (b[i] == -1) throw;
    sw(y, ry, b[i], m);
    m = b[i];
    u[m] = 1;
    for (int j = 0; j < M; ++j)
      if (BT(x[m], j) != BT(y[m], j)) {
        if (j <= i) throw;
        if (b[j] == -1) throw;
        xo(x, rx, m, b[j]);
      }
    for (int j = 0; j < n; ++j)
      if (!u[j] && (y[j] & bi)) {
        xo(y, ry, j, m);
      }
  }
  for (int i = ry.size(); i; --i) rx.push_back(ry[i - 1]);
  printf("%d\n", rx.size());
  for (int i = 0; i < rx.size(); ++i)
    printf("%d %d\n", rx[i].first, rx[i].second);
  for (int i = 0; i < rx.size(); ++i) {
    xx[rx[i].first - 1] ^= xx[rx[i].second - 1];
  }
  for (int i = 0; i < n; ++i)
    if (xx[i] != yy[i]) throw;
  return 0;
}